14
votes

how to avoid the loop to reduce the computation time of this code (one solution of my last question):

I hope to find the column vectors of A(1:3,:) whose corresponding values in M(4,:) are not part of one of the vectors of the cell X (and obviously not equal to one of these vectors). I look for a fast solution if X is very large.

M = [1007  1007  4044  1007  4044  1007  5002 5002 5002 622 622;
      552   552   300   552   300   552   431  431  431 124 124; 
     2010  2010  1113  2010  1113  2010  1100 1100 1100  88  88;
        7    12    25    15    12    30     2   10   55  32  12];

Here I take directly A:

A = [1007  4044  5002  622;
      552   300   431  124;
     2010  1113  1100   88];

A contains unique column vectors of M(1:3,:)

X = {[2 5 68 44],[2 10 55 9 17],[1 55 6 7 8 9],[32 12]};

[~, ~, subs] = unique(M(1:3,:)','rows');

A4 = accumarray(subs(:),M(4,:).',[],@(x) {x});

%// getting a mask of which columns we want
idxC(length(A4)) = false;
for ii = 1:length(A4)
    idxC(ii) = ~any(cellfun(@(x) all(ismember(A4{ii},x)), X));
end

Displaying the columns we want

out = A(:,idxC)

Results:

>> out

out =

    1007        4044
     552         300
    2010        1113

the column vector [5002;431;1100] was eliminated because [2;10;55] is contained in X{2} = [2 10 55 9 17]

the column vector [622;124;88] was eliminated because [32 12] = X{4}

Another example: with the same X

    M = [1007  4044  1007  4044  1007  5002 5002 5002 622 622  1007  1007  1007;
          552   300   552   300   552   431  431  431 124 124   552    11    11; 
         2010  1113  2010  1113  2010  1100 1100 1100  88  88  2010    20    20;
           12    25    15    12    30     2   10   55  32  12     7    12     7];

X = {[2 5 68 44],[2 10 55 9 17],[1 55 6 7 8 9],[32 12]};

A = [1007  4044  5002  622  1077;
      552   300   431  124    11;
     2010  1113  1100   88    20];

Results: (with scmg answer)

I get if A sorted according to the first row: (correct result)

out =

         1007        1007        4044
           11         552         300
           20        2010        1113

if I do not sort the matrix A, I get: (false result)

out =

        4044        5002         622
         300         431         124
        1113        1100          88

the column vector A(:,4) = [622;124;88] should be eliminated because [32 12] = X{4}.

the column vector [5002;431;1100] should be eliminated because [2;10;55] is contained in X{2} = [2 10 55 9 17]

4
Could you explan the logic of how you obtain the output? That will save us time trying to deduce it from your codeLuis Mendo
@LuisMendo: I received two answer for my question. the scmg response gives the rignt output, as in the example, but it requires a lot of computation time if X is very large. The logic developed by Ben Voigt is interesting but the output result is false, and I can not find out why! input in my question are M, A and X and output is out = A(:,idxC)bzak
@LuisMendo: I hope to find the column vectors of A(1:3,:) whose the corresponding values in M(4,:) are not part of one of the vectors of the cell X (and obviously not equal to one of these vectors). I look for a fast solution if X is very large.bzak
Just to clarify: you mean "whose corresponding values in M(4,:) are not part of the same vector of the cell X", right?Luis Mendo
@LuisMendo: yes, the same vector of the cell X.bzak

4 Answers

4
votes

The answer of Ben Voigt is great, but the line for A4i = A4{ii} is the one causing issues : the for loop doesn't work this way with column vectors :

%row vector
for i = 1:3
    disp('foo');
end

    foo
    foo
    foo

%column vector
for i = (1:3).'
    disp('foo');
end

    foo

Just try for A4i = A4{ii}.' instead and it should get your work done!

Now, if we look at the output :

A(:,idxC) =

    4044        5002
     300         431
    1113        1100

As you can see, the final result is not what we expected.

As long as unique does a kind of sort, the subs are not numbered by the order of encounter in A, but by order of encounter in C (which is sorted) :

subs =

 2
 2
 3
 2
 3
 2
 4
 4
 4
 1
 1

Therefore you should pass by the matrix given by unique rather than A to get your final output

Enter

[C, ~, subs] = unique(M(1:3,:)','rows'); 
%% rather than [~, ~, subs] = unique(M(1:3,:)','rows');

Then, to get the final output, enter

>> out = C(idxC,:).'
out =

        1007        4044
         552         300
        2010        1113
4
votes

In this case, you should not be trying to eliminate loops. The vectorization is actually hurting you badly.

In particular (giving a name to your anonymous lambda)

issubset = @(x) all(ismember(A4{ii},x))

is ridiculously inefficient, because it doesn't short-circuit. Replace that with a loop.

Same for

any(cellfun(issubset, X))

Use an approach similar to this instead:

idxC = true(size(A4));
NX = numel(X);
for ii = 1:length(A4)
    for jj = 1:NX
        xj = X{jj};
        issubset = true;
        for A4i=A4{ii}
            if ~ismember(A4i, xj)
                issubset = false;
                break;
            end;
        end;
        if issubset
            idxC(ii) = false;
            break;
        end;
    end;
end;

The two break statements, and especially the second one, trigger an early exit that potentially saves you a huge amount of computation.

3
votes

Shot #1

Listed in this section is an approach that is supposed to be a quick and direct approach to solve our case. Please note that since A is the matrix of unique columns from M considering upto the third row, it is skipped here as the input because we generate it internally with the solution code. This is maintained in the next approach/shot as well. Here's the implementation -

function out = shot1_func(M,X)

%// Get unique columns and corresponding subscripts
[unqrows, ~, subs_idx] = unique(M(1:3,:)','rows');
unqcols = unqrows.'; %//'

counts = accumarray(subs_idx(:),1); %// Counts of each unique subs_idx

%// Modify each cell of X based on their relevance with the fourth row of M
X1 = cellfun(@(x) subs_idx(ismember(M(4,:),x)),X,'Uni',0);

lensX = cellfun('length',X1); %// Cell element count of X1

Xn = vertcat(X1{:}); %// Numeric array version of X
N = max(subs_idx);   %// Number of unique subs_idx

%// Finally, get decision mask to select the correst columns from unqcols
sums = cumsum(bsxfun(@eq,Xn,1:N),1);
cumsums_at_shifts = sums(cumsum(lensX),:);

mask1 = any(bsxfun(@eq,diff(cumsums_at_shifts,[],1),counts(:).'),1); %//'
decision_mask = mask1 | cumsums_at_shifts(1,:) == counts(:).';    %//'
out = unqcols(:,~decision_mask);

return

Shot #2

The earlier mentioned approach might have a bottleneck at :

cellfun(@(x) subs_idx(ismember(M4,x)),X,'Uni',0)

So, alternatively to keep performance as a good motivation, one can separate out the whole process into two stages. The first stage could take care of cells of X that are not repeated in the fourth row of M, which could be implemented with a vectorized approach and another stage solving for the rest of X's cells with our slower cellfun based approach.

Thus, the code would bloat out a bit, but hopefully would be better with performance. The final implementation would look something like this -

%// Get unique columns and corresponding subscripts
[unqrows, ~, subs_idx] = unique(M(1:3,:)','rows')
unqcols = unqrows.' %//'
counts = accumarray(subs_idx,1);

%// Form ID array for X
lX = cellfun('length',X)
X_id = zeros(1,sum(lX))
X_id([1 cumsum(lX(1:end-1)) + 1]) = 1
X_id = cumsum(X_id)

Xr = cellfun(@(x) x(:).',X,'Uni',0); %//'# Convert to cells of row vectors
X1 = [Xr{:}]                         %// Get numeric array version

%// Detect cells that are to be processed by part1 (vectorized code)
[valid,idx1] = ismember(M(4,:),X1)
p1v = ~ismember(1:max(X_id),unique(X_id(accumarray(idx1(valid).',1)>1))) %//'

X_part1 = Xr(p1v)
X_part2 = Xr(~p1v)

%// Get decision masks from first and second passes and thus the final output
N = size(unqcols,2);
dm1 = first_pass(X_part1,M(4,:),subs_idx,counts,N)
dm2 = second_pass(X_part2,M(4,:),subs_idx,counts)
out = unqcols(:,~dm1 & ~dm2)

Associated functions -

function decision_mask = first_pass(X,M4,subs_idx,counts,N)

lensX = cellfun('length',X)'; %//'# Get X cells lengths
X1 = [X{:}];                  %// Extract cell data from X

%// Finally, get the decision mask
vals = changem(X1,subs_idx,M4) .* ismember(X1,M4);

sums = cumsum(bsxfun(@eq,vals(:),1:N),1);
cumsums_at_shifts = sums(cumsum(lensX),:);
mask1 = any(bsxfun(@eq,diff(cumsums_at_shifts,[],1),counts(:).'),1); %//'
decision_mask = mask1 | cumsums_at_shifts(1,:) == counts(:).';    %//'
return


function decision_mask = second_pass(X,M4,subs_idx,counts)

%// Modify each cell of X based on their relevance with the fourth row of M
X1 = cellfun(@(x) subs_idx(ismember(M4,x)),X,'Uni',0);

lensX = cellfun('length',X1); %// Cell element count of X1

Xn = vertcat(X1{:}); %// Numeric array version of X
N = max(subs_idx);   %// Number of unique subs_idx

%// Finally, get decision mask to select the correst columns from unqcols
sums = cumsum(bsxfun(@eq,Xn,1:N),1);
cumsums_at_shifts = sums(cumsum(lensX),:);

mask1 = any(bsxfun(@eq,diff(cumsums_at_shifts,[],1),counts(:).'),1); %//'
decision_mask = mask1 | cumsums_at_shifts(1,:) == counts(:).';       %//'

return

Verficication

This section lists code to verify the output. Here's the code to do so to verify the shot #1 code -

%// Setup inputs and output
load('matrice_data.mat');   %// Load input data
X = cellfun(@(x) unique(x).',X,'Uni',0); %// Consider X's unique elements
out = shot1_func(M,X); %// output with Shot#1 function

%// Accumulate fourth row data from M based on the uniqueness from first 3 rows
[unqrows, ~, subs] = unique(M(1:3,:)','rows');    %//'
unqcols = unqrows.';                              %//'
M4 = accumarray(subs(:),M(4,:).',[],@(x) {x});    %//'
M4 = cellfun(@(x) unique(x),M4,'Uni',0);

%// Find out cells in M4 that correspond to unique columns unqcols
[unqcols_idx,~] = find(pdist2(unqcols.',out.')==0);

%// Finally, verify output
for ii = 1:numel(unqcols_idx)
    for jj = 1:numel(X)
        if all(ismember(M4{unqcols_idx(ii)},X{jj}))
            error('Error: Wrong output!')
        end
    end
end
disp('Success!')
1
votes

Maybe you can use 2 times cellfun:

idxC = cellfun(@(a) ~any(cellfun(@(x) all(ismember(a,x)), X)), A4, 'un', 0);
idxC = cell2mat(idxC);
out = A(:,idxC)